# Finding the Jordan Canonical Form of a Classical Adjoint of a Jordan Block

Let $A$ be a size $n$ Jordan matrix with $0$ on its diagonal, that is $$A = J_n(0) = [a_{ij}] = \begin{cases} 1, &j=i+1\\ 0, &\text{elsewhere} \end{cases}$$

What is the Jordan Canonical Form of the classical adjoint of A, $\text{adj} A$?

Can we start with the fact that $A$ is singular and $A (\text{adj} A) = 0_n?$

• If the matrix is in Jordan form and is 0s on its diagonal, the last row of the matrix is all 0 and the matrix is singular, shouldn't it be? – RGS Nov 14 '16 at 9:35
• Are you talking about the adjoint? – user198504 Nov 14 '16 at 9:38
• I am talking about A. A has a row that is only 0, the bottom one. Isn't it? – RGS Nov 14 '16 at 9:38
• The first column is zero – user198504 Nov 14 '16 at 9:39
• Yup, and the last row as well. Thus $A$ is not non-singular. $A$ is singular – RGS Nov 14 '16 at 9:40

If you just start computing the classical adjoint for $n=2,3,4...$ you should notice a pattern as to what they look like.

$$adj\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & - 1\\ 0 & 0\end{pmatrix}$$

$$adj\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

$$adj \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}$$

Once you prove that this pattern holds, the Jordan Form is straightforward to compute.

• Oh cool. But dont you think that the adjoint should alternated between 1 and -1 as n varies. Also, I think the only nonzero entry will be in the topright corner, not in the buttom left corner. – user198504 Nov 16 '16 at 8:44
• You are right about the sign. The bottom left corner is definitely correct though. – Ken Duna Nov 16 '16 at 14:08
• Oh wait, I forgot to take the transpose! You were right on both counts! – Ken Duna Nov 16 '16 at 14:14